Question

Cal 1 Question (Optimization)

Answer 1

The base of the box is square, so if the sides of the base are $x$ and the box has height $y$, then the volume of the box is

$V=x^2y$

We are maximizing this subject to the constraint of surface area,

$72=\underbrace{5x^2}_{\text{glass top}}+\underbrace{4xy}_{\text{lateral faces}}+\underbrace{x^2}_{\text{base}}$
$72=6x^2+4xy$

Solving for $y$ gives

$y=\dfrac{72-6x^2}{4x}$

and substituting into the volume equation gives

$V=x^2\dfrac{72-6x^2}{4x}$
$V=18x-\dfrac32x^3$

Differentiating, we get

$V'=18-\dfrac92x^2$

which has roots at $x=\pm2$, but we omit the negative root. So $x=2$.

This means we must have

$y=\dfrac{72-6(2)^2}{4(2)}=6$